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Stels [109]
3 years ago
11

A racecar on a straight track, starting from rest*, steps on the

Physics
1 answer:
kirill115 [55]3 years ago
6 0
I think the answer is C
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Two guitarists attempt to play the same note of wavelength 6.50 cm at the same time, but one of the instruments is slightly out
PolarNik [594]

Answer:

The two value of the wavelength for the out of tune guitar is  

\lambda _2 = (6.48,6.52) \ cm

Explanation:

From the question we are told that

     The wavelength of the note is \lambda  =  6.50 \ cm = 0.065 \ m

     The difference in beat frequency is \Delta  f = 17.0 \ Hz

     

Generally the frequency of the note played by the guitar that is in tune is  

        f_1 = \frac{v_s}{\lambda}

Where v_s is the speed of sound with a constant value v_s  =  343 \ m/s

       f_1 = \frac{343}{0.0065}

      f_1 = 5276.9 \ Hz

The difference in beat is mathematically represented as

       \Delta  f =  |f_1 - f_2|

Where f_2 is the frequency of the sound from the out of tune guitar

     f_2 =f_1  \pm \Delta f

substituting values

      f_2 =f_1 + \Delta f

      f_2 = 5276.9 + 17.0  

     f_2 = 5293.9 \ Hz

The wavelength for this frequency is

      \lambda_2 = \frac{343 }{5293.9}

     \lambda_2 = 0.0648 \ m

    \lambda_2 = 6.48 \ cm

For the second value of the second frequency

     f_2 =  f_1 - \Delta f

     f_2 = 5276.9 -17

      f_2 = 5259.9 Hz

The wavelength for this frequency is

   \lambda _2 = \frac{343}{5259.9}

   \lambda _2 = 0.0652 \ m

   \lambda _2 = 6.52 \ cm

8 0
3 years ago
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